最大偶数不大于 N
给定一个数字 N，我们需要写一个程序来寻找最大的不大于 N 的数字，它的所有数字都是偶数。 例:
Input: N = 23
Output: 22
Explanation: 22 is the largest number not
greater then N which has all digits even.
Input: N = 236
Output: 228
Explanation: 228 is the largest number not
greater than N which has all digits even.
天真法:天真法是从 N 到 0 迭代，找到第一个所有数字都是偶数的数字。 以下是上述办法的实施情况:
C++
// CPP program to print the largest
// integer not greater than N with all even digits
#include <bits/stdc++.h>
using namespace std;
// function to check if all digits
// are even of a given number
int checkDigits(int n)
{
// iterate for all digits
while (n) {
if ((n % 10) % 2) // if digit is odd
return 0;
n /= 10;
}
// all digits are even
return 1;
}
// function to return the largest number
// with all digits even
int largestNumber(int n)
{
// iterate till we find a
// number with all digits even
for (int i = n;; i--)
if (checkDigits(i))
return i;
}
// Driver Code
int main()
{
int N = 23;
cout << largestNumber(N);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to print the largest
// integer not greater than N with
// all even digits
import java .io.*;
public class GFG {
// function to check if all digits
// are even of a given number
static int checkDigits(int n)
{
// iterate for all digits
while (n > 0)
{
// if digit is odd
if (((n % 10) % 2) > 0)
return 0;
n /= 10;
}
// all digits are even
return 1;
}
// function to return the largest
// number with all digits even
static int largestNumber(int n)
{
// iterate till we find a
// number with all digits even
for (int i = n;; i--)
if (checkDigits(i) > 0)
return i;
}
// Driver Code
static public void main (String[] args)
{
int N = 23;
System.out.println(largestNumber(N));
}
}
// This code is contributed by vt_m.
Python 3
# Python3 program to print the largest
# integer not greater than N with
# all even digits
# function to check if all digits
# are even of a given number
def checkDigits(n):
# iterate for all digits
while (n>0):
# if digit is odd
if ((n % 10) % 2):
return False;
n =int(n/10);
# all digits are even
return True;
# function to return the
# largest number with
# all digits even
def largestNumber(n):
# Iterate till we find a
# number with all digits even
for i in range(n,-1,-1):
if (checkDigits(i)):
return i;
# Driver Code
N = 23;
print(largestNumber(N));
# This code is contributed by mits
C
// C# program to print the largest
// integer not greater than N with
// all even digits
using System;
public class GFG {
// function to check if all digits
// are even of a given number
static int checkDigits(int n)
{
// iterate for all digits
while (n > 0)
{
// if digit is odd
if (((n % 10) % 2) > 0)
return 0;
n /= 10;
}
// all digits are even
return 1;
}
// function to return the largest
// number with all digits even
static int largestNumber(int n)
{
// iterate till we find a
// number with all digits even
for (int i = n;; i--)
if (checkDigits(i) > 0)
return i;
}
// Driver Code
static public void Main ()
{
int N = 23;
Console.WriteLine(largestNumber(N));
}
}
// This code is contributed by aunj_67.
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to print the largest
// integer not greater than N with
// all even digits
// function to check if all digits
// are even of a given number
function checkDigits($n)
{
// iterate for all digits
while ($n)
{
// if digit is odd
if (($n % 10) % 2)
return 0;
$n /= 10;
}
// all digits are even
return 1;
}
// function to return the
// largest number with
// all digits even
function largestNumber($n)
{
// Iterate till we find a
// number with all digits even
for ($i = $n; ; $i--)
if (checkDigits($i))
return $i;
}
// Driver Code
$N = 23;
echo(largestNumber($N));
// This code is contributed by Ajit.
?>
java 描述语言
<script>
// javascript program to print the largest
// integer not greater than N with
// all even digits
// function to check if all digits
// are even of a given number
function checkDigits(n)
{
// iterate for all digits
while (n > 0)
{
// if digit is odd
if (((n % 10) % 2) > 0)
return 0;
n = parseInt(n/ 10);
}
// all digits are even
return 1;
}
// function to return the largest
// number with all digits even
function largestNumber(n)
{
// iterate till we find a
// number with all digits even
for (i = n;; i--)
if (checkDigits(i) > 0)
return i;
}
// Driver Code
var N = 23;
document.write(largestNumber(N));
// This code is contributed by shikhasingrajput
</script>
输出:
22
时间复杂度: O(N) 高效方法:我们可以通过将 N 中的第一个奇数数字减 1，然后用最大的偶数数字(即 8)替换该奇数数字右侧的所有数字来获得所需的数字。例如，如果 N = 24578，那么 X = 24488。在某些情况下，这种方法可以创建一个前导 0，在这种情况下，我们可以简单地丢弃前导 0。例如，如果 N = 1334，那么 X = 0888。所以，我们的答案将是 X = 888。如果 N 中没有奇数，那么 N 就是数字本身。 以下是上述方法的实施:
C++
// CPP program to print the largest
// integer not greater than N with all even digits
#include <bits/stdc++.h>
using namespace std;
// function to return the largest number
// with all digits even
int largestNumber(int n)
{
string s = "";
int duplicate = n;
// convert the number to a string for
// easy operations
while (n) {
s = char(n % 10 + 48) + s;
n /= 10;
}
int index = -1;
// find first odd digit
for (int i = 0; i < s.length(); i++) {
if ((s[i] - '0') % 2 & 1) {
index = i;
break;
}
}
// if no digit, then N is the answer
if (index == -1)
return duplicate;
int num = 0;
// till first odd digit, add all even numbers
for (int i = 0; i < index; i++)
num = num * 10 + (s[i] - '0');
// decrease 1 from the odd digit
num = num * 10 + (s[index] - '0' - 1);
// add 0 in the rest of the digits
for (int i = index + 1; i < s.length(); i++)
num = num * 10 + 8;
return num;
}
// Driver Code
int main()
{
int N = 24578;
cout << largestNumber(N);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to print the largest
// integer not greater than N with all even digits
class GFG
{
// function to return the largest number
// with all digits even
static int largestNumber(int n)
{
String s = "";
int duplicate = n;
// convert the number to a string for
// easy operations
while (n > 0)
{
s = (char)(n % 10 + 48) + s;
n /= 10;
}
int index = -1;
// find first odd digit
for (int i = 0; i < s.length(); i++)
{
if ((((int)(s.charAt(i) - '0') % 2) & 1) > 0)
{
index = i;
break;
}
}
// if no digit, then N is the answer
if (index == -1)
return duplicate;
int num = 0;
// till first odd digit, add all even numbers
for (int i = 0; i < index; i++)
num = num * 10 + (int)(s.charAt(i) - '0');
// decrease 1 from the odd digit
num = num * 10 + ((int)s.charAt(index) - (int)('0') - 1);
// add 0 in the rest of the digits
for (int i = index + 1; i < s.length(); i++)
num = num * 10 + 8;
return num;
}
// Driver Code
public static void main (String[] args)
{
int N = 24578;
System.out.println(largestNumber(N));
}
}
// This code is contributed by mits
Python 3
# Python3 program to print the largest
# integer not greater than N with
# all even digits
import math as mt
# function to return the largest
# number with all digits even
def largestNumber(n):
s = ""
duplicate = n
# convert the number to a string
# for easy operations
while (n > 0):
s = chr(n % 10 + 48) + s
n = n // 10
index = -1
# find first odd digit
for i in range(len(s)):
if ((ord(s[i]) - ord('0')) % 2 & 1):
index = i
break
# if no digit, then N is the answer
if (index == -1):
return duplicate
num = 0
# till first odd digit, add all
# even numbers
for i in range(index):
num = num * 10 + (ord(s[i]) - ord('0'))
# decrease 1 from the odd digit
num = num * 10 + (ord(s[index]) -
ord('0') - 1)
# add 0 in the rest of the digits
for i in range(index+1,len(s)):
num = num * 10 + 8
return num
# Driver Code
N = 24578
print(largestNumber(N))
# This code is contributed
# by Mohit kumar 29
C
// C# program to print the largest
// integer not greater than N with all even digits
using System;
class GFG
{
// function to return the largest number
// with all digits even
static int largestNumber(int n)
{
string s = "";
int duplicate = n;
// convert the number to a string for
// easy operations
while (n > 0)
{
s = (char)(n % 10 + 48) + s;
n /= 10;
}
int index = -1;
// find first odd digit
for (int i = 0; i < s.Length; i++)
{
if ((((int)(s[i] - '0') % 2) & 1) > 0)
{
index = i;
break;
}
}
// if no digit, then N is the answer
if (index == -1)
return duplicate;
int num = 0;
// till first odd digit, add all even numbers
for (int i = 0; i < index; i++)
num = num * 10 + (int)(s[i] - '0');
// decrease 1 from the odd digit
num = num * 10 + ((int)s[index] - (int)('0') - 1);
// add 0 in the rest of the digits
for (int i = index + 1; i < s.Length; i++)
num = num * 10 + 8;
return num;
}
// Driver Code
static void Main()
{
int N = 24578;
Console.WriteLine(largestNumber(N));
}
}
// This code is contributed by mits
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to print the largest
// integer not greater than N with all even digits
// function to return the largest number
// with all digits even
function largestNumber($n)
{
$s = "";
$duplicate = $n;
// convert the number to a string for
// easy operations
while ($n)
{
$s = chr($n % 10 + 48).$s;
$n =(int)($n/10);
}
$index = -1;
// find first odd digit
for ($i = 0; $i < strlen($s); $i++)
{
if (ord($s[$i] - '0') % 2 & 1)
{
$index = $i;
break;
}
}
// if no digit, then N is the answer
if ($index == -1)
return $duplicate;
$num = 0;
// till first odd digit, add all even numbers
for ($i = 0; $i < $index; $i++)
$num = $num * 10 + (ord($s[$i]) - ord('0'));
// decrease 1 from the odd digit
$num = $num * 10 + ((ord($s[$i]) - ord('0')) - 1);
// add 0 in the rest of the digits
for ($i = $index + 1; $i < strlen($s); $i++)
$num = $num * 10 + 8;
return $num;
}
// Driver Code
$N = 24578;
echo largestNumber($N);
// This code is contributed by mits
?>
java 描述语言
<script>
// javascript program to print the largest
// integer not greater than N with all even digits
// function to return the largest number
// with all digits even
function largestNumber(n)
{
var s = "";
var duplicate = n;
// convert the number to a string for
// easy operations
while (n > 0)
{
s = String.fromCharCode(n % 10 + 48) + s;
n = parseInt(n/10);
}
var index = -1;
// find first odd digit
for (i = 0; i < s.length; i++)
{
if ((((s.charAt(i).charCodeAt(0) - '0'.charCodeAt(0)) % 2) & 1) > 0)
{
index = i;
break;
}
}
// if no digit, then N is the answer
if (index == -1)
return duplicate;
var num = 0;
// till first odd digit, add all even numbers
for (i = 0; i < index; i++)
num = num * 10 + (s.charAt(i).charCodeAt(0) - '0'.charCodeAt(0));
// decrease 1 from the odd digit
num = num * 10 + (s.charAt(index).charCodeAt(0) - ('0').charCodeAt(0) - 1);
// add 0 in the rest of the digits
for (i = index + 1; i < s.length; i++)
num = num * 10 + 8;
return num;
}
// Driver Code
var N = 24578;
document.write(largestNumber(N));
// This code contributed by shikhasingrajput
</script>
输出:
24488
时间复杂度: O(M)，其中 M 为位数
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